Question

A 75.0-liter canister contains 15.82 moles of argon at a pressure of 546.8 kilopascals. What is the temperature of the canister?

Answer 1

Answer: 312 K.

Step-by-step explanation:

Answer 2

• 312 K

Step-by-step explanation:

1) Data:

a) V = 75.0 liter

b) n = 15.82 mol

c) p = 546.8 kPa

d) T = ?

2) Formula:

• Ideal gas equation: pV = nRT

Where:

• n = number of moles
• V = volume
• p = absolute pressure
• T = absolute temperature
• R = Universal Gas constat: 8.314 kPa - liter / K-mol

3) Solution:

a) Solve the equation for T:

• T = pV / (nR)

b) Substitute and compute:

• T = 546.8 kPa × 75.0 l iter / (15.82 mol × 8.314 kPa-liter/K-mol) = 312 K

(since the volume is expressed with 3 significant figures, the answer must show also 3 significant figures)